I am wondering if there is any mathematical (or physical, besides the fact that classical quantum mechanics uses complex numbers) justification for why the complexified (1,3) Clifford algebra is used in Dirac's equation. A (the?) key point of special relativity is that spacetime is a real 4-d vector space with an inner product of signature (1,3). But by complexifying the signature becomes irrelevant-- all complex Clifford algebras in a given dimension are isomorphic where for real Clifford algebras, even signatures (p,q) and (q,p) are not isomorphic in general (as a side question: can there be a physical significance to this fact? or do only the spin group and the even subalgebra of the Clifford algebra, which are the same for (p,q) and (q,p), matter? I only hear about spinor bundles and spin structures, never pinor bundles or pin structures).

Thanks and I hope this isn't too physicsy of a question!


3 Answers 3


This problem was investigated by Cecile DeWitt-Morette et al. This is a review article describing the role of Pin groups in Physics. This article includes also a historical survey and a comprehensive list of references about Pin groups.

Becides the fact that the Clifford algebras Cl(3,1) and Cl(1,3) are nonisomorphic, there is the question if the two types can be experimentally distinguished. The article gives a positive in the neutrinoless double beta decay experiment where it was observed that the neutrino has to be a Pin(1,3) particle.

The article also describes solutions of the Dirac equation in topologically nontrivial spaces where the vacuum expectations of the Fermi currents are different in the two types of groups. (Thus one might conclude that this difference can be used to get information about the space-time topology).

The article also refers to Choquet-Bruhat, DeWitt-Morette and Dillard-Bleick's book about obstructions to the construction of Pin bundles.

  • $\begingroup$ +1 for the reference, which I didn't know. $\endgroup$ Mar 14, 2010 at 18:14
  • $\begingroup$ Thanks for the article-- this is the type of thing I was looking for. $\endgroup$ Mar 14, 2010 at 21:06

I am slightly confused by this question. The fact that one can formulate the Dirac equation in either (3,1) or (1,3) signature, which have non-isomorphic Clifford algebras and hence Clifford modules of different type (real for (1,3) and quaternionic for (3,1) in my naming conventions), does not mean that one is complexifying the Clifford algebra in formulating the Dirac equation.

Unitarity of the time evolution -- a physical requirement independent of choices -- requires that $i D$, where $D$ is the Dirac operator, be hermitian, and in turn this forces a certain hermiticity condition on the "gamma" matrices, in essence choosing a real form of the complex Clifford algebra. It is the spinor representation which can be taken to be complex since after all wave functions live in the tensor product of the Clifford module with a complex Hilbert space. This is not the same thing as complexifying the Clifford algebra, though.

So to summarise, in the Dirac equation the Clifford algebra is real, but the pinor representation can be taken to be complex.

  • $\begingroup$ In Dirac theory multiplication by i is allowed, so we can pass from (1,3) to (3,1) (I guess this comes from what you're saying about the wave fns being in the tensor prod of the real (s)pinor space with complex Hilbert space). But i has no counterpart in either Cl(1,3) or Cl(3,1) (both have trivial centers). The essence of my question is that by using C (which is necessary for classical QM) don't we disrespect the real Clifford structure (which is natural from Lorentz geometry). Is this necessary? It seems like quaternions could give a middle ground. $\endgroup$ Mar 14, 2010 at 21:03

The main point in signature (1,3) is that the corresponding Laplacian is the wave operator, whereas in (0,4) you have the standard Laplacian (up to overall sign changes, but that doesn't matter). Hence your differential equations are of a different type, even though your functions may be complex-valued.
Hence there is a big difference between solutions of
$\Delta u=0$ for $u:M\rightarrow \mathbb{C}$ (elliptic) and
$\square u=0$ for $u:M\rightarrow \mathbb{C}$ (hyperbolic)
Now translate this to your Dirac equation (being some sort of square root of the above), where the functions you consider are the same (sections of isomorphic spinor bundles), but the Dirac operators $\gamma^i \nabla_i$ are different.


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